Question

Find two consecutive positive integers whose sum of squares is 613

Answer 1

Hey Mate!

let the consecutive numbers be x and x + 1

Sum of squares = 613

${x}^{2} + {(x + 1)}^{2} = 613 \\ \\ {x}^{2} + {x}^{2} + 2x + 1 = 613 \\ \\ 2 {x}^{2} + 2x - 612 = 0 \\ \\ 2( {x}^{2} + x - 306) = 0 \\ \\ {x}^{2} + x - 306 = 0$

${x}^{2} + 18x - 17x - 30 6 = 0 \\ \\ x(x + 18) = 0 \: \: \: \: x(x - 17) = 0 \\ \\ x = - 17 \: \: \: \: x = 18$
Take x = 17. Therefore , the numbers are 17 and 18


HOPE THIS HELPS U...

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

Smaller of the two consecutive positive integers be p.

Hence,

Second integer will be p + 1.

Situation would be like :-

p² + (p + 1)² = 613

p² + p² + 2 × p × 1 + 1² = 613

p² + p² + 2p + 1 - 613 = 0

2p² + 2p - 612 = 0

2(p² + p - 306) = 0

p² + p - 306 = 0

$\textbf{\underline{Splitting\;the\;middle\;Term}}$

p² + 18p - 17p - 143 = 0

p(p + 18) - 17 (p + 18) = 0

(p - 17) (p + 18) = 0

p = 17

p = - 18

Given :-

$\textbf{\underline{Odd\;Positive\;Integer}}$

Hence,

p = 17.

$\textbf{\underline{Therefore\;we\;get}}$

$\Large{\boxed{\sf\:{Two\;consecutive\;odd\;integers\;are\;17 \;and\;18.}}}$